How many grams of al(oh)3 (molar mass = 78.0 g/mol) can be produced from the reaction of 48.6 ml of .15 m koh with excess al2(so
1 answer:
Taking into account the reaction stoichiometry, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
Al₂(SO₄)₃ + 6 KOH → 2 Al(OH)₃ + 3 K₂SO₄
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al₂(SO₄)₃: 1 mole
- KOH: 6 moles
- Al(OH)₃: 2 moles
- K₂SO₄: 3 moles
The molar mass of the compounds is:
- Al₂(SO₄)₃: 342 g/mole
- KOH: 56.1 g/mole
- Al(OH)₃: 78 g/mole
- K₂SO₄: 174.2 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
- KOH: 6 moles ×56.1 g/mole= 336.6 grams
- Al(OH)₃: 2 moles ×78 g/mole= 156 grams
- K₂SO₄: 3 moles ×174.2 g/mole= 522.6 grams
Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume:
Molarity= number of moles÷ volume
<h3>Mass of Al(OH)₃ formed</h3>In firts place, you know that 48.6 mL (0.0486 L) of 0.15 M KOH react with excess Al₂(SO₄)₃.
Replacing in the definition, you can calculate the amount of moles of KOH that react:
0.15 M= number of moles÷ 0.0486 L
Solving:
0.15 M× 0.0486 L= number of moles
<u><em>0.00729 moles= number of moles</em></u>
Then, 0.00729 moles of KOH react with excess Al₂(SO₄)₃. So, following rule of three can be applied: if by reaction stoichiometry 6 moles of KOH form 156 grams of Al(OH)₃, 0.00729 moles of KOH form how much mass of Na₂SO₄?
<u><em>mass of Al(OH)₃= 0.18954 grams</em></u>
Then, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.
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the reaction stoichiometry:
molarity:
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